Let A = 0 00 G|5 0 0 0 , u = 1|5 15 10 - 25 and v= g d h Define T: R³ R³ by T(x) = Ax. Find T(u) and T(v). →>>
Let A = 0 00 G|5 0 0 0 , u = 1|5 15 10 - 25 and v= g d h Define T: R³ R³ by T(x) = Ax. Find T(u) and T(v). →>>
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Linear Algebra: Transformations**
Let \( A =
\begin{bmatrix}
\frac{1}{5} & 0 & 0 \\
0 & \frac{1}{5} & 0 \\
0 & 0 & \frac{1}{5} \\
\end{bmatrix}
\), \( u =
\begin{bmatrix}
15 \\
10 \\
-25 \\
\end{bmatrix}
\), and \( v =
\begin{bmatrix}
g \\
d \\
h \\
\end{bmatrix}
\).
Define \( T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 \) by \( T(x) = Ax \).
Find \( T(u) \) and \( T(v) \).
**Calculation**
\[ T(u) = \_\_\_\_\_\_\_\_\_\_\_\_\_ \]
(Simplify your answer. Use integers or fractions for any numbers in the expression.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2F5c14fe25-cfb6-4bd3-8d3a-f9f38ff3bbac%2Faydb3_processed.png&w=3840&q=75)
Transcribed Image Text:**Linear Algebra: Transformations**
Let \( A =
\begin{bmatrix}
\frac{1}{5} & 0 & 0 \\
0 & \frac{1}{5} & 0 \\
0 & 0 & \frac{1}{5} \\
\end{bmatrix}
\), \( u =
\begin{bmatrix}
15 \\
10 \\
-25 \\
\end{bmatrix}
\), and \( v =
\begin{bmatrix}
g \\
d \\
h \\
\end{bmatrix}
\).
Define \( T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 \) by \( T(x) = Ax \).
Find \( T(u) \) and \( T(v) \).
**Calculation**
\[ T(u) = \_\_\_\_\_\_\_\_\_\_\_\_\_ \]
(Simplify your answer. Use integers or fractions for any numbers in the expression.)
![**Determine if** \( v \) **is in the set spanned by the columns of** \( B \).
\[
B = \begin{bmatrix}
3 & -3 & 2 \\
9 & -7 & 9 \\
6 & 0 & 15 \\
9 & -5 & 16 \\
\end{bmatrix}
\]
\[
v = \begin{bmatrix}
-12 \\
-20 \\
24 \\
-4 \\
\end{bmatrix}
\]
---
**Choose the correct answer below and, if necessary, fill in the answer box(es) to complete your choice.**
- \( \textcircled{A} \) Vector \( v \) is not in the set spanned by the columns of \( B \) because the columns of \( B \), \( b_1 \), \( b_2 \), and \( b_3 \) are linearly independent.
- \( \textcircled{B} \) Vector \( v \) is in the set spanned by the columns of \( B \) because the columns of \( B \) span \(\mathbb{R}^4 \).
- \( \textcircled{C} \) Vector \( v \) is not in the set spanned by the columns of \( B \) because the reduced echelon form of the matrix formed by writing \( B \) with a fourth column equal to \( v \) is \(\square\).
- \( \textcircled{D} \) Vector \( v \) is in the set spanned by the columns of \( B \) because \(\square b_1 + \square b_2 + \square b_3 = v \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2F5c14fe25-cfb6-4bd3-8d3a-f9f38ff3bbac%2F9eslqxs_processed.png&w=3840&q=75)
Transcribed Image Text:**Determine if** \( v \) **is in the set spanned by the columns of** \( B \).
\[
B = \begin{bmatrix}
3 & -3 & 2 \\
9 & -7 & 9 \\
6 & 0 & 15 \\
9 & -5 & 16 \\
\end{bmatrix}
\]
\[
v = \begin{bmatrix}
-12 \\
-20 \\
24 \\
-4 \\
\end{bmatrix}
\]
---
**Choose the correct answer below and, if necessary, fill in the answer box(es) to complete your choice.**
- \( \textcircled{A} \) Vector \( v \) is not in the set spanned by the columns of \( B \) because the columns of \( B \), \( b_1 \), \( b_2 \), and \( b_3 \) are linearly independent.
- \( \textcircled{B} \) Vector \( v \) is in the set spanned by the columns of \( B \) because the columns of \( B \) span \(\mathbb{R}^4 \).
- \( \textcircled{C} \) Vector \( v \) is not in the set spanned by the columns of \( B \) because the reduced echelon form of the matrix formed by writing \( B \) with a fourth column equal to \( v \) is \(\square\).
- \( \textcircled{D} \) Vector \( v \) is in the set spanned by the columns of \( B \) because \(\square b_1 + \square b_2 + \square b_3 = v \).
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